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10 June, 06:05

What is the change in entropy when 0.210 mol of potassium freezes at 63.7°C? (ΔHfus = 2.39 kJ/mol) kJ/K

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  1. 10 June, 08:55
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    1.75 J/K.

    Explanation:

    Using Gibbs free energy equation,

    ΔG = ΔH - TΔS

    Where,

    ΔG = free energy

    ΔH = enthalpy change

    T = absolute temperature

    ΔS = entropy change

    Note: ΔG of fusion = 0

    Therefore,

    ΔH = TΔS,

    ΔS = ΔH/T

    Given:

    T = °C + 273.15

    = 273.15 + 69

    = 342.15 K

    n = 0.25 mol.

    ΔH = 2.39 kJ/mol

    = 2.39 / 342.15

    = 6.99 J/Kmol

    ΔS = ΔSmolar * n

    = 6.99 * 0.25

    = 1.75 J/K.
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